The weight of 1000 children average 50 kg and the standard deviation i

1.	The weight of 1000 children average 50 kg and the standard deviation is 5 kg. what is the z-score corresponds to 55?Solution:
Answer» Answer: E( X ˉ )=μ X ˉ =μ=50 kg and a variance of {\sigma^2 \over n}={5^2 \over 30}={5 \over 6} n σ 2 = 30 5 2 = 6 5 The standard deviation of the sampling distribution \sigma_{\bar{X}}={5 \over \sqrt{30}}={\sqrt{30}\over 6}σ X ˉ = 30 5 = 6 30 2) P(48<\bar{X}<53)=P\bigg({48-50 \over 6/\sqrt{30}} X ˉ <53)=P( 6/ 30 48−50 6/ 30 53−50 )= =P\bigg(Z<{\sqrt{30} \over 2}\bigg)-\bigg(1-P\bigg(Z<{\sqrt{30} \over 3}\bigg)\bigg)\approx=P(Z< 2 30 )−(1−P(Z< 3 30 ))≈ \approx0.9969-0.0339=0.9630≈0.9969−0.0339=0.9630 The number of sample that fall betweeen 48 and 53 kilograms 100\cdot0.9630=96.3\approx96100⋅0.9630=96.3≈96 Explanation: I hope it is helpful for you

The weight of 1000 children average 50 kg and the standard deviation is 5 kg. what is the z-score corresponds to 55?Solution:

Answer»

Answer:

)=μ

=μ=50 kg

and a variance of

{\sigma^2 \over n}={5^2 \over 30}={5 \over 6}

The standard deviation of the sampling distribution

\sigma_{\bar{X}}={5 \over \sqrt{30}}={\sqrt{30}\over 6}σ

P(48<\bar{X}<53)=P\bigg({48-50 \over 6/\sqrt{30}}

<53)=P(

48−50

53−50

=P\bigg(Z<{\sqrt{30} \over 2}\bigg)-\bigg(1-P\bigg(Z<{\sqrt{30} \over 3}\bigg)\bigg)\approx=P(Z<

)−(1−P(Z<

))≈

\approx0.9969-0.0339=0.9630≈0.9969−0.0339=0.9630

The number of sample that fall betweeen 48 and 53 kilograms

100\cdot0.9630=96.3\approx96100⋅0.9630=96.3≈96

Explanation:

I hope it is helpful for you